Theories

The Frequency Theory

The proportion of times it might have
happened in the past that it actually did, e.g.,
p("I get run over crossing 38th street after class") =

(Number of times people got run over crossing 38th st.)
-------------------------------------------------------
(Number of times people crossed 38th st.)

But why this denominator? why the numerator?

Other theories

The Logical Theory

The proportion of all possible exchangeable (i.e., equally
likely) worlds which entail our proposition of interest being
true. E.g. Playing cards.

But how often can we apply this in the real world?

The Personal Theory

Probability is a subjective judgment based
on all of the knowledge and beliefs you have. There is no
objectively perfect way to determine the `correct' probability.
Reasonable people can disagree, because they have different
evidence available to them.

Rules of coherence

p(A) + p(not A) = 1

(not A is called the "complement" of A)

rain

not rain

Additivity

Mutually Exclusive

Propositions A and B are "mutually exclusive" if they
cannot both be true at the same time.

I.e., if one of the propositions is true, that "excludes" the
possibility of the other being true: the two propositions "mutually
exclude" each other.

When propositions A and B are mutually exclusive:
p(A or B) = p(A) + p(B)

e.g. p(yes) = p(female-yes or male-yes) = p(male-yes) + p(female-yes)

male-yes

female-yes

Where does additivity come from?

From betting. The expected value of a bet on an event is its
probability times the amount to win. (Later we'll see that this works
for "expected utility.")

For example, the expected value of "$10 if a coin comes up heads"
is $5.

What is the EV of "$10 if a coin comes up heads twice (in 2 flips)"?

EV is (roughly) the average value if the bet were repeated.

EV and additivity

We can bet on two events at once.

The value of a bet on one event should not change when you
break it into two events.

For example, the expected value of (and willingness to pay for)

$10 if "Red card"

should be the same as:

$10 if "Heart"and $10 if "Diamond"

Example, continued

Suppose you are willing to pay or accept $4 for $10 if "Red card"

$2.50 for $10 if "Heart"

$2.50 for $10 if "Diamond"

You have a ticket for the first bet. I buy it from you for $4.
You are now ahead by $4.

Example, continued

Suppose you are willing to pay or accept $4 for $10 if "Red card"

$2.50 for $10 if "Heart"

$2.50 for $10 if "Diamond"

You have a ticket for the first bet. I buy it from you for $4.
You are now ahead by $4.

Then I sell you tickets for the second and third bets for $2.50
each. You are now behind by $1.

Example, continued

Suppose you are willing to pay or accept $4 for $10 if "Red card"

$2.50 for $10 if "Heart"

$2.50 for $10 if "Diamond"

You have a ticket for the first bet. I buy it from you for $4.
You are now ahead by $4.

Then I sell you tickets for the second and third bets for $2.50
each. You are now behind by $1.

Then I point out that the second and third bets together are the same as the
first bet, so you are willing to trade those two tickets for a
ticket for the first bet. You are still behind by $1, and we start
over....

Conditional Probability Defined

The conditional probability of proposition A given
proposition B is the probability that we would assign to A
if we knew that B were true, that is, the probability of A
conditional on B being true. We write p(A|B) or p(A/B). This does not
mean "divided by".

For example, what is the probability that Obama wins if Perry is the
nominee?

Multiplication Rule - p(A and B)

The Conditional Probability Rule is: p(A | B) = p(A and B) / p(B)

(This time the / does mean "divided by".)

In other words: p(A and B) / p(B) = p(A | B)

or:
p (A and B) = p(A | B) × p(B), the multiplication rule.

p (Obama-win and Perry-nom) = p(Obama-win | Perry-nom) × p(Perry-nom)

Conditioned assessment (Kleinmuntz et al., 1996)

Estimate the probability of some event E.

Estimate the probability of E given some other event F, p(E|F)

Estimate p(E|not-F)

Estimate p(F)

Compute E' as p(F)×p(E|F) + p(not-F)×p(E|not-F)

Independent Propositions

Two propositions A and B are independent if believing that A is
true does not change your belief about whether B is true. (The reverse
always holds.)

Are "It is raining" and "Melissa is mowing the lawn" independent?

Are "The first car to pass us will be a Dodge" and "The second car
to pass us will be a BMW" independent?

When A and B are independent, then the multiplication rule can be
simplified, because p(A | B) = p(A). Hence: p(A and B) = p(A) × p(B).